What is the derivative of y=(t+3)in(t) ?

The derivative of y=(t+3)in(t) is i(2nt+3n)

What is the first derivative of y=(t+3)in(t) ?

The first derivative of y=(t+3)in(t) is i(2nt+3n)

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derivative of

y = (t + 3) in (t)

i (2 n t + 3 n)

Solution steps

\frac{d}{d t} ((t + 3) in t)

Simplify

(t + 3) in t : i (n t^{2} + 3 n t)

= \frac{d}{d t} (i (n t^{2} + 3 n t))

Take the constant out:

(a \cdot f)^{'} = a \cdot f^{'}

= i \frac{d}{d t} (n t^{2} + 3 n t)

Apply the Sum/Difference Rule:

(f \pm g)^{'} = f^{'} \pm g^{'}

= i (\frac{d}{d t} (n t^{2}) + \frac{d}{d t} (3 n t))

\frac{d}{d t} (n t^{2}) = 2 n t

\frac{d}{d t} (3 n t) = 3 n

= i (2 n t + 3 n)

\frac{d}{d x} (\frac{x ^{2} - 4 x + 2}{x + 3})

\int 1 - 1 d x

\int \frac{1}{4 + x ^{2}} d x

\int x^{3} x + 10 d x

x \to 0 lim (tan (x^{2}))

What is the derivative of y=(t+3)in(t) ?
The derivative of y=(t+3)in(t) is i(2nt+3n)
What is the first derivative of y=(t+3)in(t) ?
The first derivative of y=(t+3)in(t) is i(2nt+3n)